Problem: $f\,^{\prime}(x)=12e^x$ and $f(4)=-16+12e^4$. $f(0) = $
Explanation: Finding $f(x)$ We have $f'(x)=12e^x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (12e^x)\,dx \\\\ & = {12e^x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(4)=-16+12e^4$. Here's what we get when we plug in $4$ : $\begin{aligned}f(4)&={12e^{4}} {+ C} \end{aligned}$ We are given that this must equal $-16+12e^4$ : $-16+12e^4 = {12e^{4}} {+ C}$ Solving the equation gives us ${C=-16}$. Finding $f(0)$ Now, we have that $f(x)={12e^x} {-16}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=12e^0 - 16\\\\ &=-4 \end{aligned}$ The answer $f(0) = -4$